X=61600

Đúng thì tk mình nha

\(1709+x-1175=62134\)

\(1709+x=62134+1175\)

\(1709+x=63309\)

\(x=63309-1709\)

\(x=61600\)

a, ( 75 x 56 ) : 15                          1175 : 25 + 1325: 25 + 466 x 25

= (75: 15) x 56                         = ( 1175 + 1325) : 25 + 466 x 25

= 3 x 56                                   =     2500 : 25 + 466 x25

= 168                                         = 100 + 466 x 25

b, ( 80  x125 x10) : 25               =   25 x 4 + 466 x 25

= ( 80 x 10) x ( 125:25)            = 25 x ( 4 + 466)

= 800 x 5                                  = 25 x 470

= 4 000                                     =    11750

d, ( 750 x 360 x 55) : ( 15 x 12 x 110)

=   ( 75 x 360 x 550 ) : ( 15 x 12 x 110)

= (75 : 15 ) x ( 360 : 12) x ( 550 : 110)

= 5 x 30 x 5

= 25 x 30

= 750 

a) (75 x 56) : 15

= 4200 : 15

= 280

b) (80 x 125 x 10) : 25

= 100 000 : 25

= 4000

c) 1175 : 25 + 1325 : 25 + 466 x 25

= (1175 : 25) + (1325 : 25) + (466 x 25)

= 47 + 53 + 11 650

= 11 750

d) (750 x 360 x 55) : (15 x 12 x 110)

= 14 850 000 : 19 800

= 750

\(a,\frac{x+1}{x-2}-\frac{x-1}{x+2}=\frac{2\left(x^2+2\right)}{x^2-4}\)

\(\Leftrightarrow\frac{\left(x+1\right)\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}-\frac{\left(x-1\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\frac{2x^2+4}{\left(x-2\right)\left(x+2\right)}\)

\(\Rightarrow x^2+2x+x+2-\left(x^2-2x-x+2\right)=2x^2+4\)

\(\Leftrightarrow x^2+3x+2-x^2+2x+x-2=2x^2+4\)

\(\Leftrightarrow6x=2x^2+4\)

\(\Leftrightarrow2x^2+4-6x=0\)

\(\Leftrightarrow2x^2+4-6x=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x+3=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=1\\x=-3\end{cases}}\)

\(b,\frac{2x+1}{x-1}=\frac{5\left(x-1\right)}{x+1}\)

\(\Leftrightarrow\left(2x+1\right)\left(x+1\right)=5\left(x-1\right)\left(x-1\right)\)

\(\Leftrightarrow2x^2+2x+x+1=5\left(x^2-2x+1\right)\)

\(\Leftrightarrow2x^2+3x+1=5x^2-10x+5\)

\(\Leftrightarrow5x^2-2x^2-10x-3x+5-1=0\)

\(\Leftrightarrow3x^2-13x+4=0\)

\(\Leftrightarrow\left(x-4\right)\left(x-\frac{1}{3}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-4=0\\x-\frac{1}{3}=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=4\\x=\frac{1}{3}\end{cases}}}\)

a: \(\Leftrightarrow\left(x-5\right)\left(x+1\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\\x=1\end{matrix}\right.\)

d: \(\Leftrightarrow\left(x+3\right)\left(x^2-4x+5\right)=0\)

\(\Leftrightarrow x+3=0\)

hay x=-3

e) Ta có: \(E=\left(3x+2\right)\left(3x-5\right)\left(x-1\right)\left(9x+10\right)+24x^2\)

\(=\left(9x^2-15x+6x-10\right)\left(9x^2+10x-9x-10\right)+24x^2\)

\(=\left(9x^2-10-9x\right)\left(9x^2-10+x\right)+24x^2\)

\(=\left(9x^2-10\right)^2-8x\left(9x^2-10\right)-9x^2+24x^2\)

\(=\left(9x^2-10\right)^2-8x\left(9x^2-10\right)+15x^2\)

\(=\left(9x^2-10\right)^2-3x\left(9x^2-10\right)-5x\left(9x^2-10\right)+15x^2\)

\(=\left(9x^2-10\right)\left(9x^2-3x-10\right)-5x\left(9x^2-10-3x\right)\)

\(=\left(9x^2-3x-10\right)\left(9x^2-5x-10\right)\)

`x^2=3`

`=>x=\sqrt{3}\or\x=-\sqrt{3}`

`x^2=36`

`<=>x^2=(+-6)^2`

`<=>x=+-6`

`x^2=25`

`<=>x^2=(+-5)^2`

`<=>x=+-5`

`2x^2+(-20)=55`

`<=>2x^2-20=55`

`<=>2x^2=75`

`<=>x^2=75/2`

`<=>x=+-\sqrt{75/2}`

`2(x-1)^2+5^0=9`

`<=>2(x-1)^2+1=9`

`<=>2(x-1)^2=8`

`<=>(x-1)^2=4`

`<=>x-1=2\or\x-1=-2`

`<=>x=3\or\x=-1`

hộ nốt câu cuối ;-; :

-(x+1)²  - 5 = 2.(-3).5

<=> - (x+1)² = -25

<=> (x+1)² = 25

<=> (x+1)² - 5² = 0

<=> (x+1 + 5).(x + 1 - 5) = 0 (hằng đẳng thức)

<=> th1 : x + 6 = 0 <=> x = -6

<=> th2: x -4 = 0 <=> x = -4

vậy tập nghiệm của pt trên là: S = {-6,-4}

a) Thực hiện rút gọn VT = -2x – 64

Giải phương trình -2x – 64 = 0 thu được x = -32.

b) Thực hiện rút gọn VT = -62 x +12

Giải phương trình -62x + 12 = -50 thu được x = 1.

a: \(A=x^2+y^2=\left(x+y\right)^2-2xy=15^2-2\cdot50=115\)

c: \(x-y=\sqrt{\left(x+y\right)^2-4xy}=\sqrt{15^2-4\cdot50}=5\)

\(C=x^2-y^2=\left(x+y\right)\left(x-y\right)=15\cdot5=75\)